A 3-digit number ABC, on multiplication with D gives 37DD where A, B, C and D are different non-zero digits. What is the value of A + B + C ?

(UPSC CSAT – 2023)

1. 18
2. 16
3. 15
4. Can’t be determined due to insufficient data.

Solution: Ans.(1)

ABC × D = 37DD,

When D multiply by C, it should be D as unit digit, therefore D is a single digit number that can give itself by multiplying any other number, so it can be 2, 4, 5 or 8.

If D is 2, then C should be 6 which gives unit digit as 2 and 1 carry forward, but it can not make again 2 by multiplying it to B.

If D is 5, then C should be any odd number, which gives unit digit 5 and some carry forward, but next it also can not make again 5.

If D is 8, then C should be 6 which gives 8 as unit digit and 4 carry forward. And B should be 3 which gives 28 (=24+4) but next any value of A can not make 37.

If D is 4, then C should be 6 which gives unit digit 4 and 2 carry forward. And B should be 3 which gives 14 (=12+2), and next A should be 9 which gives 37 (=36+1).

Hence  A + B + C = 9 + 3 + 6 = 18.